What is the integral of 1/(x(x^6+8)) ?

The integral of 1/(x(x^6+8)) is 1/6 (3/4 ln(x)-1/8 ln|x^6+8|)+C

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integral of 1/(x(x^6+8))

\int \frac{1}{x ( x ^{6} + 8 )} d x

\frac{1}{6} (\frac{3}{4} ln (x) - \frac{1}{8} ln x^{6} + 8) + C

Solution steps

\int \frac{1}{x ( x ^{6} + 8 )} d x

Apply u-substitution

: \int \frac{1}{6 u ( u + 8 )} d u

= \int \frac{1}{6 u ( u + 8 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u ( u + 8 )} d u

Take the partial fraction of

\frac{1}{u ( u + 8 )} : \frac{1}{8 u} - \frac{1}{8 ( u + 8 )}

= \frac{1}{6} \cdot \int \frac{1}{8 u} - \frac{1}{8 ( u + 8 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int \frac{1}{8 u} d u - \int \frac{1}{8 ( u + 8 )} d u)

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

\int \frac{1}{8 ( u + 8 )} d u = \frac{1}{8} ln ∣ u + 8 ∣

= \frac{1}{6} (\frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣)

Substitute back

u = x^{6}

= \frac{1}{6} (\frac{1}{8} ln x^{6} - \frac{1}{8} ln x^{6} + 8)

\frac{1}{8} ln x^{6} = \frac{3}{4} ln (x)

= \frac{1}{6} (\frac{3}{4} ln (x) - \frac{1}{8} ln x^{6} + 8)

Add a constant to the solution

= \frac{1}{6} (\frac{3}{4} ln (x) - \frac{1}{8} ln x^{6} + 8) + C

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What is the integral of 1/(x(x^6+8)) ?
The integral of 1/(x(x^6+8)) is 1/6 (3/4 ln(x)-1/8 ln|x^6+8|)+C